Does uniform convergence in $L_p$ implies uniform convergence in probability?

Question

Suppose we have a sequence of random variables $X_n(\theta)$ that depends on some parameter $\theta$ which varies over a compact set $\Theta$. Suppose we have

\begin{equation}\tag{1}\label{eq1} \sup_{\theta \in \Theta} \Vert X_n(\theta) - X(\theta)\Vert = o(1), \end{equation} is it at all possible to claim for any $\epsilon >0$, $$\mathbb{P} (\sup_{\theta \in \Theta}\vert X_n(\theta) - X(\theta)\vert > \epsilon) \to 0.\tag{2}\label{eq2}$$

It is not immediately obvious to me because the supremum in \eqref{eq1} is taken outside of the norm, not inside. By this I mean if $\Vert \sup_{\theta \in \Theta} \vert X_n(\theta) - X(\theta) \vert \Vert = o(1)$, then sure we can claim that $\sup_{\theta \in \Theta} \vert X_n(\theta) - X(\theta) \vert \to 0$ in probability. Can we assert a stronger mode of convergence than \eqref{eq2} with \eqref{eq1}?

Answer 1

Take $\Omega = \Theta = [0, 1]$, $X_n(\theta)(\omega) = \mathbb I_{[\theta - 1 / n, \theta + 1/n]}(\omega)$ and $X \equiv 0$.

Then $\|X_n(\theta) - X(\theta)\| \leq 2/n$ for any $\theta$, thus we have uniform convergence in $L_p$

However, $\mathbb P(\exists \theta: |X_n(\theta) - X(\theta)| > 1/2) = 1$, thus we don't have uniform convergence in probability.